Question 6.S.P.3: Assume that the panel considered in Sample Problem 1 is actu...

From Sample Problem 1, the midplane strains and curvatures are:

ε°_{xx} = 0  μm/m,      \kappa _{xx} = -0.50   rad/m

ε°_{yy} = -1300  μm/m,      \kappa _{yy} = 0.40   rad/m

γ°_{xy} = 900  μrad,      \kappa _{xy} = -0.20   rad/m

To determine ply interface positions, first note that the total laminate thickness is:

A total of nine ply interface positions must be determined because there are eight plies in the laminate. Following the numbering scheme discussed in Sec. 4 and referring to Fig. 11, ply interface positions are:

z_{0}=-t/2=-(0.001 m)=-0.000500 m
z_{1}=z_{0}+t_{1}=-0.000500 m+0.000125 m=-0.000375 m
z_{2}=z_{1}+t_{2}=-0.000375 m+0.000125 m=-0.000250 m
z_{3}=z_{2}+t_{3}=-0.000250 m+0.000125 m=-0.000125 m
z_{4}=z_{3}+t_{4}=-.000125 m+0.000125 m=0.000000 m
z_{5}=z_{4}+t_{5}=0.000000 m+0.000125 m=0.000125 m
z_{6}=z_{5}+t_{6}=0.000125 m+0.000125 m=0.000250 m
z_{7}=z_{6}+t_{7}=0.000250 m+0.000125 m=0.000375 m
z_{8}=z_{7}+t_{7}=0.000375 m+0.000125 m=0.000500 m

Strain Calculations. Strains are calculated using Eq. (12), and can be determined at any through-thickness position. Usually, strains of greatest interest are those induced at the ply interface locations. For example, strains present at the outer surface of ply 1 (i.e., strains present at z_{o}=-0.000500 m) are:

\left\{\begin{matrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \gamma _{xy} \end{matrix} \right\} = \left\{\begin{matrix} \varepsilon° _{xx} \\ \varepsilon° _{yy} \\ \gamma° _{xy} \end{matrix} \right\} + z \left\{\begin{matrix} \kappa _{xx} \\ \kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\}       (12)

Similarly, strains present at the interface between plies 1 and 2 (i.e., strains present at z_{1}=-0.000375 m) are:

Strains present at all remaining interfaces are calculated in exactly the same fashion. Strains calculated at all ply interfaces are summarized in Table 1 and are plotted in Fig. 13. Note that all three strain components (ε_{xx}, ε_{yy}, and γ_{xy})are predicted to be linearly distributed through the plate thickness. This linear distribution is a direct consequence of the Kirchhoff hypothesis, which is a good approximation as long as the plate is ‘‘thin.’’ In fact, identical strain distributions would be predicted for any thin plate subjected to the midplane strains and curvatures specified in Sample Problem 1. For example, we would predict the identical strains if an aluminum plate were under consideration rather than a laminated composite plate.

The strains listed in Table 1 and plotted in Fig. 13 are referenced to the global x–y coordinate system. As will be seen, knowledge of ply strains referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the strain tensor from one coordinate system to another was reviewed in Chap. 2 and, in particular, strains can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (44) of Chap. 2. In practice, strains are usually calculated at both the ‘‘top’’ and ‘‘bottom’’ interface for each ply. Example calculations for plies 1 and 2 are listed below: Ply 1. Because θ_{1}=0°, the x–y and 1–2 coordinate systems are coincident, and therefore the description of the strain tensor is identical in both coor-dinate systems. This can be confirmed through application of Eq. (44) Chap. 2:

Top interface:

\left\{\begin{matrix} \varepsilon _{x^{\prime }x^{\prime }} \\ \varepsilon _{y^{\prime } y^{\prime }} \\ \frac{\gamma _{x^{\prime }y^{\prime }}}{2} \end{matrix} \right\}= \left[\begin{matrix} \cos ^{2}(\theta) & \sin ^{2}(\theta) & 2 \cos (\theta) \sin(\theta) \\ \sin ^{2} (\theta) & \cos ^{2} (\theta)&-2 \cos (\theta) \sin (\theta) \\ – \cos (\theta) \sin (\theta) & \cos (\theta) \sin (\theta)& \cos^{2} (\theta)- \sin^{2}(\theta) \end{matrix} \right] \left\{\begin{matrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \frac{\gamma _{xy}}{2} \end{matrix} \right\}       (44)

Bottom interface:

Ply 2. In this case, θ_{2}=30° and consequently the description of strain in the x–y and 1–2 coordinate systems differs substantially. Applying Eq. (44) of Chap. 2, we have:
Top interface:

Bottom interface:

Ply strains referenced to the local 1–2 coordinate systems at all interface locations are summarized in Table 2 and plotted in Fig. 14. Comparing Figs. 12 and 13, it is apparent that the through-thickness strain distributions no longer appear linear or continuous when referenced to the 1–2 coordinate system. This is of course illusionary, in the sense that strains appear to be dis-continuous only because the coordinate system used to describe the through-thickness strain is varied from one ply to the next.

Stress Calculations. Because strains are now known at all ply interface positions, we can calculate stresses at these locations using Eq. (30) of Chap. 5, with ΔT=ΔM=0. During these calculations, we will require the trans-formed reduced stiffness matrix for each ply. Using graphite–epoxy material properties from Table 2 of Chap. 3 and Eqs. (11) and (31) of Chap. 5, we find:

M_{xx}= \int_{-t/2}^{t/2}{\left\{z \overline{Q}_{11} \varepsilon _{xx}^{o}+ z \overline{Q}_{12} \varepsilon _{yy}^{o} + z \overline{Q}_{16} \gamma _{xx}^{o} + z^{2} \overline{Q}_{11} \kappa _{xx} + z^{2} \overline{Q}_{12} \kappa _{yy} + z^{2} \overline{Q}_{16} \kappa _{xy}\right\} }dz        (30)

Q_{11} = \frac{E_{11}^{2}}{E_{11} ν_{12}^{2} E_{22}}    Q_{12} = Q_{21}= \frac{ν_{12}E_{11}E_{22}}{E_{11} ν_{12}^{2} E_{22}}\\ Q_{22}= \frac{E_{11}E_{22}}{E_{11} ν_{12}^{2} E_{22}}    Q_{16} = G_{12}     (11)

\overline{Q}_{11} = Q_{11} cos^{4} \theta + 2(Q_{12} + 2Q_{66})cos^{2}\theta   sin^{2}\theta + Q_{22} sin^{4}\theta \\\overline{Q}_{12} =\overline{Q}_{21} = Q_{12} ( cos^{4} \theta + sin^{4}\theta) +(Q_{11} +Q_{22} -4 Q_{66})cos^{2}\theta   sin^{2}\theta \\ \overline{Q}_{16} =\overline{Q}_{61} = (Q_{11}- Q_{12} -2Q_{66})cos^{3}\theta   sin\theta- (Q_{22}-Q_{12}-2Q_{66})cos\theta   sin^{3}\theta\\ \overline{Q}_{22} = Q_{11}sin^{4} \theta + 2(Q_{12} + 2Q_{66})cos^{2}\theta   sin^{2}\theta + Q_{22} cos^{4}\theta \\ \overline{Q}_{26} =\overline{Q}_{62} = (Q_{11} – Q_{12} – 2Q_{66})cos\theta   sin^{3}\theta -(Q_{22} -Q_{12} – 2Q_{66})cos^{3}\theta   sin\theta \\ \overline{Q}_{66} = (Q_{11} +Q_{22} -2Q_{12} -2Q_{66}) cos^{2}\theta   sin^{2}\theta +Q_{66} ( cos^{4} \theta + sin^{4}\theta)      (31)

For 0° plies:

For 30° plies:

For 90° plies:

For -30° plies:

Stresses present at the outer surface of ply 1 (i.e., strains present at z_{0}=-0.000500 m) can now be calculated:

To calculate stresses at the interface between plies 1 and 2 (i.e., at z_{1} =-0.000375 m), we must specify whether we are interested in the stresses within ply 1 or ply 2. That is, according to our idealized model, a ply interface is treated as a plane of discontinuity in material properties. Ply 1 ‘‘ends’’ at z =z_{1}^{(-)}, whereas ply 2 ‘‘begins’’ at z=z_{1}^{(+)}. Hence, the stresses within ply 1 at z =z_{1}^{(-) } are:

The stresses within ply 2 (a 30° ply) at z=z_{1}^{(+)} are:

Stresses are calculated at all remaining ply interfaces in exactly the same fashion. Ply interface stresses are summarized in Table 3 and are plotted in Fig. 15. Obviously, stresses are not linearly distributed through the thickness of the laminate, even when referenced to the global x–y coordinate system. In general, all stress components exhibit a sudden discontinuous change at all ply interface positions. The abrupt change in stresses at ply interfaces is due to the discontinuous change in the [\overline{Q}] matrix from one ply to the next. In turn, the discontinuous change in [\overline{Q}] occurs because the fiber angle (in general) changes from one ply to the next. Indeed, in this example problem, the same fiber angle occurs in only two adjacent plies (namely, plies 4 and 5, both of which have a fiber angle of -30°), and inspection of Fig. 15 shows that the interface between plies 4 and 5 is the only interface for which the stresses do not change abruptly.

It has been mentioned that the linear strain distributions shown in Fig. 13 would be the same for any thin plate, regardless of the material the plate is made of. The same statement cannot be made for stress distributions. In general, through-thickness stress distributions for isotropic plates (e.g., an isotropic aluminum plate) are linear and continuous, unless high nonlinear stresses occur, in which case the stress distribution may not be linear but will nevertheless be continuous. In contrast, the stress distributions in laminated composite plates are usually discontinuous. The only conditions under which a linear and continuous stress distribution is encountered is when: (a) the laminate is subjected to elastic stress/strain levels, and (b) when the [\overline{Q}] matrix does not vary from one ply to the next (i.e., for unidirectional laminates in which the fiber angle does not vary from one ply to the next).

Knowledge of ply stresses referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the stress tensor from one coordinate system to another was reviewed in Chap. 2 and, in particular, stresses can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (20) of Chap. 2. Typically, stresses are calculated at both the ‘‘top’’ and ‘‘bottom’’ interfaces for all plies. For example, rotation of the ply stresses that exist within ply 2 at the interface between plies 1 and 2 (i.e., at z=z_{1} = -0.375 mm) proceeds as follows:

\left\{\begin{matrix} \sigma _{x^{\prime }x^{\prime }} \\ \sigma _{y^{\prime } y^{\prime }} \\ \tau _{x^{\prime }y^{\prime }} \end{matrix} \right\}= \left[\begin{matrix} \cos ^{2}(\theta) & \sin ^{2}(\theta) & 2 \cos (\theta) \sin(\theta) \\ \sin ^{2} (\theta) & \cos ^{2} (\theta)&-2 \cos (\theta) \sin (\theta) \\ – \cos (\theta) \sin (\theta) & \cos (\theta) \sin (\theta)& \cos^{2} (\theta)- \sin^{2}(\theta) \end{matrix} \right]\\\times \left\{\begin{matrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau _{xy} \end{matrix} \right\}         (20)

Ply interface stresses referenced to local 1–2 coordinate systems are summarized in Table 4 and plotted in Fig. 16. Once again, stresses are not linearly distributed through the thickness of the laminate, and instead exhibit a sudden discontinuous change at all ply interface positions.
Stress invariants can be used to confirm that the ply stresses referenced to the x–y coordinate system and listed in Table 3 are equivalent to the ply stresses referenced to the 1–2 coordinate system, as listed in Table 4. The concept of ‘‘stress invariants’’ was discussed in Chap. 2. The stress invariants for the case of plane stress are given by Eq. (22) of Chap. 2, repeated here for convenience:

First stress invariant =\Theta = \sigma _{xx} + \sigma _{yy}

Second stress invariant = \Phi = \sigma _{xx} \sigma _{yy} – \tau _{xy}^{2}     (repeated) (2.22)

Third stress invariant = \Psi  = 0

For plane stress conditions, the third stress invariant always equals zero, and so \Psi   cannot be used to evaluate whether two plane stress states are equivalent. The first and second stress invariants, \Theta  and \Phi, respectively, have been calculated using the ply stress components referenced to both the x–y and 1–2 coordinate systems. Values calculated for \Theta  and \Phi are included in the last two columns of both Tables 3 and 4. Identical values are obtained in all cases, indicating the equivalence of the ply stress states described using the two different coordinate systems.